Find the probability of getting a heart on the first draw and a club on the second draw while drawing 2 cards successively and without replacement

Question

I am currently having some difficulty with a card probability question (I guess combinatorics will haunt me for the whole life...). The question is to find the probability of getting a heart on the first draw and a club on the second draw while drawing 2 cards successively and without replacement from the standard poker deck. There are 2 approaches that I came with but they seem to produce different results.

1. Logically, I believe the answer to be $$ P(C|H)=\frac{\binom{13}{1}}{\binom{52}{1}}*\frac{\binom{13}{1}}{\binom{39}{1}}=\frac{1}{12}=0.083 $$ However, I can't formally prove that it is the right approach, so I tried using the conditional probability formula.

2. $$ P(C|H)=\frac{P(C\cap H)}{P(H)}=\frac{\binom{13}{1}*\binom{13}{1}}{\binom{52}{2}}/\frac{\binom{13}{1}}{\binom{52}{1}}=\frac{2}{51}=0.0392 $$

One(or even 2) of my answers are clearly wrong, so I would be glad if someone could tell me where my mistake is and how to fix one(or both) approaches.

Answer 1

The straightforward way to answer this question is to see that the probability of a heart on the first draw is $13/52$, or, equivalently, $1/4$.

Then there are $51$ cards left, so the probability of a spade the next time is $13/51$.

Multiply.

(I haven't addressed your question about what's wrong with your proposed approaches. You don't say why you think either might be right.)